MATH 308 Lecture 31

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Begin Exam 4 content

Eigenvalues and Eigenvectors

$A={\begin{bmatrix}2&-3\\1&-2\end{bmatrix}}$

$\lambda \in \{-1,1\}$

$\lambda =1:{\vec {v_{1}}}=\left\langle 3,1\right\rangle$

$\lambda =-1:{\vec {v_{2}}}=\left\langle 1,1\right\rangle$

Find the General solution of $X'(t)=A\,X$ and describe the behavior as $t\to \infty$.

We need to find two linearly independent solutions and take a linear combination of them.

Let $X_{2}(t)=f_{2}(t)\,V_{2}$, where $f_{2}$ is any scalar function, and $V_{2}$ is the eigenvector for $\lambda =-1$.

So for our diff eq, $X_{2}'(t)=A\,X_{2}(t)$

We get

${\begin{aligned}f_{2}'(t)\,V_{2}&=f_{2}(t)\,(A\,V_{2}(t))\\&=f_{2}(t)\,(-1)\,V_{2}\end{aligned}}$

Recall that $(A-(-1)\,I)\,V_{2}=0$, so $A\,V_{2}=(-1)\,V_{2}$. That's where the second form came from.

If we plug in $V_{2}=\left\langle 1,1\right\rangle$, we get

${\begin{bmatrix}f_{2}'(t)\\f_{2}'(t)\end{bmatrix}}={\begin{bmatrix}-f_{2}(t)\\-f_{2}(t)\end{bmatrix}}$

Thus $f_{2}'(t)=-f_{2}(t)$, and the solution to this differential equation is $f_{2}(t)=C\,\mathrm {e} ^{-t}$

Putting this back in Matrix form, we get

$X_{2}(t)=C\,\mathrm {e} ^{-t}\,{\begin{bmatrix}1\\1\end{bmatrix}}$

A similar process for $V_{1}$ gives

$f_{1}(t)=C\,\mathrm {e} ^{t}$

And

$X_{1}(t)=C\,\mathrm {e} ^{t}\,{\begin{bmatrix}3\\1\end{bmatrix}}$

Therefore, the general solution is

${\begin{bmatrix}x(t)\\y(t)\end{bmatrix}}=c_{1}\,\mathrm {e} ^{t}\,{\begin{bmatrix}3\\1\end{bmatrix}}+c_{2}\,\mathrm {e} ^{-t}\,{\begin{bmatrix}1\\1\end{bmatrix}}$